Volume 6 No. 1 pp. 56–67 (2013)c IEJG
VISUAL DISTINGUISHABILITY OF SEGMENTS
ARP ´´ AD KURUSA (Communicated by Levent KULA)
Abstract. By considering the equioptics of segments in a plane we find some answers to the question that “if the measures of the visual angles of two seg- ments are equal at some points in the plane do they coincide?”.
1. Introduction
Theshadow picture[4] of a compact convex set B is defined at each pointP ∈ R2\ B as the angle measure νB(P) of the so called visual angle that B subtends at P. The point P and the set B are usually called the source and the object of shadow pictureνB(P). The shadow picture is often called visual angle(sometimes this makes some confusion) orpoint projection[2]. The functionνBis usually called visual angle function.
The central problem in this subject is to show such setS of sources and set O of objects thatO 3 B 7→νB S is injective. There are a number of such results in the literature [3], [4], [5], [6], [7], [10] etc.
In this article we consider the distinguishability of segments by investigating first their equioptics in detail.
2. Equioptics of segments
Theequiopticof two compact convex sets is the set of those points, where their shadow pictures are equal.1 Thecompopticof two compact convex sets is the set of those points, where the sum of their shadow pictures isπ (these shadow pictures are said to besupplementary).
Lemma 2.1. The equioptic of two different segments is a union of subarcs of two cubic algebraic curves. The remaining subarcs of these two cubic algebraic curves constitute the compoptic of those segments.
Date: Received: April 12, 2012 and Accepted: February 11, 2013.
2000Mathematics Subject Classification. 0052, 0054.
Key words and phrases. visual angle, shadow picture, masking function.
Supported by the T ´AMOP-4.2.2.A-11/1/KONV-2012-0073 project.
1As the shadow picture of a segment is not well defined at the endpoints of the segment, we think of it as equal to any angle in [0, π].
56
Proof. Denote the different segments by AB and CD. Let m be a unit normal vector of the plane. If the segment AB subtends the same angleϕ at X as CD does, then
(−−→
XA×−−→
XB)h−−→
XC,−−→
XDi=ε(X) (−−→
XC×−−→
XD)h−−→
XA,−−→
XBi,
where
ε(X) =
(+1, if|−−→
XC×−−→
XD|−−→
XA×−−→
XB=|−−→
XA×−−→
XB|−−→
XC×−−→
XD,
−1, otherwise.
Obviously, ε(X) is constant ±1 in every quadrants of the straight lines AB and CD, and it changes sign if and only ifX moves over one of the straight linesAB andCD.
Fix an originOand letx=−−→
OX,a=−→
OA,b=−−→
OB,c=−−→
OC andd=−−→
OD. Then
−−→XA×−−→
XB= (a−x)×(b−x) =x×(a−b) +a×b, h−−→
XA,−−→
XBi=ha−x,b−xi=|x|2− hx,a+bi+ha,bi, and
−−→XC×−−→
XD= (c−x)×(d−x) =x×(c−d) +c×d, h−−→
XC,−−→
XDi=hc−x,d−xi=|x|2− hx,c+di+hc,di, therefore
|x|2hx,(a−b)×mi+ha×b,mi|x|2− hx,(a−b)×mihx,c+di+
+hx,hc,di((a−b)×m)−ha×b,mi(c+d)i+ha×b,mihc,di
= ¯ε(x)(|x|2hx,(c−d)×mi+hc×d,mi|x|2−hx,(c−d)×mihx,a+bi+ (2.1)
+hx,ha,bi((c−d)×m)−hc×d,mi(a+b)i+hc×d,miha,bi), where ¯ε(x) = ¯ε(−−→
OX) :=ε(X). This proves that the equioptic is a union of arcs of two cubic algebraic curves. Since the equation for the compoptic is the same as (2.1), but with−¯ε, the second statement of the theorem is also proven.
As x=0 is a solution of equation (2.1) if and only ifha×b,mihc,di=hc× d,miha,bi, we infer that the endpoints A, B, C and D, and, if they exist, the intersection pointsM0:=AB∩CD, M+ :=AC∩BD andM− :=AD∩BC are on the equioptic.
The cubic algebraic curves
|x|2hx,(a−b)×mi+ha×b,mi|x|2− hx,(a−b)×mihx,c+di+
+hx,hc,di((a−b)×m)−ha×b,mi(c+d)i+ha×b,mihc,di
=±(|x|2hx,(c−d)×mi+hc×d,mi|x|2−hx,(c−d)×mihx,a+bi+ (2.2)
+hx,ha,bi((c−d)×m)−hc×d,mi(a+b)i+hc×d,miha,bi), are calledApollonian curves [11] and are denoted byA± according to the sign of the right side of (2.2). The equioptic and the compoptic are assembled from the arcs of these Apollonian curves. Both Apollonian curves pass through the points
A, B, C, D and M0, but, in general, each one passes only the point M± with its index2.
A B
C
D
Observations 2.2. The equioptic and compoptic subarcs of an Apollonian curve follow each other alternately and one takes over at those points, where the Apol- lonian curve intersects one of the straight lines of the segments. Each Apollonian curve has a straight line asymptotic to it at its both “ends”.
If an Apollonian curve intersects a straight line in more than three points, then by B´ezout’s theorem3 it contains that straight line as a component, and therefore it is reducible.
Theorem 2.3. If an Apollonian curve is reducible, then
(a)it is a nondegenerate circle and a straight line through its centre, or (b)it is a degenerate circle and a straight line, or
(c) it is a nondegenerate equilateral hyperbola and the line at infinity, or (d)it is two perpendicular straight lines and the line at infinity.
Theorem 2.3 is clearly stated in [11, p. 358. l. -7] without proof, but it has a transparent proof in [12, Section 6] and a more detailed proof in [9, Section 2].
Theorem 2.4. Let A be an Apollonian curve of the different nondegenerate seg- mentsAB andCD. Then the followings hold.
(1) If A contains an open arc Hˆ of a nondegenerate equilateral hyperbola H, then the segments are opposite sides of a parallelogram and are separated by the branches of Hif and only if they are the shorter edges in the paral- lelogram.
(2) If A contains an open arc Cˆof a circle C, then it also contains a straight line` passing through the centre ofC.
(3) If Acontains an open segment`ˆof a straight line`, then the segments are (3.1) the reflections of each other with respect to`, or
(3.2) opposite sides of a kite symmetric with respect to`, or (3.3) adjoining sides of a kite symmetric with respect to`, or (3.4) placed on`.
Proof. In this proof we regard the common ideal point of two parallel straight lines as their intersection at the infinity.
(1) As A contains an arc ˆH of the nondegenerate equilateral hyperbolaH, by B´ezout’s theorem it also contains H, hence it is reducible. Then by Theorem 2.3
2Only the coincidence of the indices ofA±andM±is invariant, because changing the order of the endpoints of one of the segment changes both indices.
3Roughly speaking B´ezout’s theorem [1] states that if two algebraic curves have more common points than the product of their degrees, then they have a common component.
the Apollonian cubic is the equilateral hyperbolaHand the straight line`∞at the infinity.
As the straight line`∞is contained byA, the straight linesABandCDare the opposite sides of a parallelogram, say it is ABCD. Further, the pairs{A, B} and {C, D}are on the same branch of the hyperbolaHif and only if the segmentsAB orCD subtend acute visual angle at the centre of the parallelogramABCD.
(2) AsAcontains an arc ˆC of the circleC, by B´ezout’s theorem it also contains C, hence it is reducible. Then by Theorem 2.3Ais the circleC and a straight line
`through its centre.
(3) AsAcontains the open segment ˆ`of the straight line`, by B´ezout’s theorem it also contains`, hence it is reducible. Then by Theorem 2.3 the Apollonian cubic can be of four types, that we are considering one-by-one.
Observe first thatAcontains the pointsA, B, C, D, M0and one of the pointsM+
andM−, and second that the segments subtend angles close to zero at the points of`near infinity, therefore they subtend the same angle at those points, hence they have projections of equal length onto any straight line`⊥ perpendicular to`.
(a) In this caseAis the union of the straight line` and a nondegenerate circle C centred on`. We consider subcases.
(a1) Assume thatA, B, C, D∈ Cand all these points are different. ThenM0∈` and M+ ∈` orM− ∈`. Suppose that M0 is exterior toC. As ABCD is a cyclic quadrilateral in C, M0AD4 is similar to M0CB4. Also the segments AB and CDare equal length chords ofC, hence by lettingx:=d(A, B),b:=d(M0, B) and c:=d(M0, C) we get (b+x)b= (c+x)c, that is (b−c)(b+c+x) = 0. Thusb=c. If M0is interior toC, then bothM±are exterior toC. AssumeM+∈`. AsABCDis a cyclic quadrilateral inC,M+CD4is similar toM+BA4. Also the segmentsAC andBDare equal length chords ofC, hence by lettingy:=d(A, C),a:=d(M+, A) andd:=d(M+, D) we get (a+y)a= (d+y)d, that is (a−d)(a+d+y) = 0. Thus a=d. (⇒(3.1))
(a2) Assume that A, B, C, D∈ C and B =C. The segments AB and CD are equal length chords of C, and they also have equal length projection on `⊥, hence they have also equal length projection on `, and therefore BA and CD close the same (undirected) angle to `. Thus, the point B = C is on the bisector of AD, which passes through the centre O of C, hence OBA∠ = OCD∠. This implies OB=`, hence the chordAD ofC is perpendicular to`, thusB =C∈`and Ais
the reflection ofDonto`. (⇒(3.3))
(a3) AssumeA, B ∈`. Then also C, D ∈`, and an easy calculation with (2.1) shows thatCis the Apollonian circle of such different length segmentsABandCD that neither one contains the other. (⇒(3.4))
(a4) AssumeA, D∈`,A=DandB∈ C \`. Then alsoC∈ C \`. The segments ABand CDsubtend equal or supplementary angles at the projectionB⊥ of B to
`, which isπ/2, soBC is perpendicular to`. Since the projections of the segments are also of equal length,`is the bisector ofBC. (⇒(3.3))
(a5) AssumeA, D∈`,A6=DandB∈ C \`. Then alsoC∈ C \`. The segments ABandCDhave equal length projection on any straight line`⊥perpendicular to`, thereforeB and Care in the same distance from `. If B=C, an easy calculation with (2.1) shows that the other component of the Apollonian that contains the straight line is a degenerate circle with centre inB=C, thereforeB 6=C, and the same reasoning as in (a4) implies that` is the bisector ofBC. (⇒(3.2))
(a6) AssumeA, C∈`andB∈ C \`. Then alsoD∈ C \`, and a simple exchanging of the name ofC and D leads us to (a4) or (a5) according to whether A=C or A6=C.
(b) In this case A is the union of the straight line ` and a degenerate circle C={P} 6⊂`.
(b1) If no points ofA, B, C, Dare inC, then an easy calculation with (2.1) shows that the other component of the Apollonian that contains`is nondegenerate circle or a straight line, therefore this case can not happen.
(b2) IfA, D∈`andB ∈ C then alsoC∈ C, and thereforeA6=D. (⇒(3.3))
(b3) IfA, C∈` andB∈ C then alsoD∈ C, therefore this is just a renaming of (b2). (⇒(3.3))
A
B C
D A
B C
D
(c) This case can not occur, because the straight line`is not at the infinity.
(d) In this caseAis the union of the straight line `and straight line`⊥ that is perpendicular to`.
Observe that the segments subtend angles close to zero at the points of`and`⊥ near infinity, therefore the segments subtend the same angle at those points, hence they have projections of equal length onto any of the straight lines ` or `⊥, and hence they have the same length.
(d1) Assume AB ⊂ `. Then also CD ⊂ `, and considering the angles the segments subtend at points of `⊥ near `∩`⊥ shows that the segments are the reflection of each other with respect to`⊥. (⇒(3.4))
(d2) AssumeAB⊂`⊥. Then alsoCD⊂`⊥, and by following the reasoning in (d1), we get that the segments are the reflection of each other with respect to `.
(⇒(3.1))
(d3) AssumeA, D∈`, A=D and B∈`⊥\`. Then alsoC∈`⊥\`, and since B6=Cthe pointCis the reflection ofB with respect to`. Then, as we saw earlier, Ais`⊥ and the circle through the pointsA, B, C that contradicts the assumption, therefore this case can not happen.
(d4) AssumeA, D∈`, A6=D andB ∈`⊥\`. Then alsoC ∈`⊥\`, and as in (d3) we conclude thatB6=C. Since the segments have equal length corresponding projections, B and C can not be on the same side of `, thus B and C are the reflections of each other with respect to`, and in the same way we deduce thatA andDare the reflections of each other with respect to`⊥. This proves thatABDC is a square. (⇒(3.2))
The proof is now completed.
3. Distinguishability of segments
A setP of points in the plane is calledset of injectivityif the coincidence of any two segments follows from the equality of the measures of their respective visual angles at every point ofP. Obviously, if a set is covered by an Apollonian curve, then it is not a set of injectivity. In this section we show sets of injectivity.
Denote the cubic polynomials on the left- and right-hand side of (2.2) byf and g, respectively, and observe thatf2=g2satisfied exactly by the points of the two Apollonian curve. Since the degree of the polynomial equationf2=g2 is 6, it has at most 6+22
= 28 independent coefficients, hence there can be chosen 29 points so that no polynomial equation of the formf2=g2can have them all as solutions.
Thus, there are sets of injectivity that contain only 29 points, and the problem arises to
(3.1) determine the minimal cardinality of a set of injectivity.
The following theorem proves that the minimal cardinality of a set of injectivity can not be more than 16.
Theorem 3.1. LetIbe the intersection point of the straight lines`1and`2that are not orthogonal to each other. Let P1, P2, P3, P4, P5, P6 ∈ `1 and Q1, Q2, Q3, Q4, Q5, Q6 ∈`2 be different points outside I and finally let R1, R2, R3
collinear points outside of `1∪`2∪`⊥1 ∪`⊥2, where the straight lines `⊥1, `⊥2 are perpendicular to`1 and `2, respectively, and pass throughI. If the nondegenerate segmentsABandCDhave equal or supplementary shadow pictures at every points of {I, P1, . . . , P6, Q1, . . . , Q6, R1, R2, R3}, then they coincide.
Proof. Assume thatABandCDare different.
As`1 has 7 common points with the Apollonian curves of ABand CD, one of the Apollonian curve, say A−, has at least 4 points common with `1, hence by B´ezout’s theorem`1 is a component ofA+.
By Theorem 2.3 we see, that outside of`1the Apollonian curveA−may intersect
`2 in at most 2 points, henceA+ must intersect `2 in at least 4 points. Thus by B´ezout’s theorem`2 is a component ofA+.
Thus both Apollonian curvesA± are of type (3) in our Theorem 2.4, and there- fore we have to investigate the pairs of the cases (3.1)–(3.4).
IfA−is of (3.1) andA+is of (3.1), then the rotation given by the reflections with respect to`1and`2is by angle 2πaroundI, hence`1=`2, which is a contradiction.
IfA− is of (3.1) andA+ is of (3.2), then the kite K mentioned in (3.2) is also symmetric to`1, hence`1is the middle line ofK, henceKis a square, and therefore A− is`1 and the circumscribed circle, A+ is `2 and the straight line of the other diagonal ofK.
IfA− is of (3.1) andA+ is of (3.3), then the kite K mentioned in (3.3) is also symmetric to`1, hence`1and`2are the diagonals ofK, that implies`1⊥`2, which is a contradiction.
IfA− is of (3.1) andA+ is of (3.4), then`1⊥`2, which is a contradiction.
IfA− is of (3.2) andA+ is of (3.2), then`1and`2are the diagonals of the kite K mentioned in (3.2), that implies`1⊥`2, which is a contradiction.
If A− is of (3.2) andA+ is of (3.3), we have a contradiction, as the segments can not be opposite and adjoining sides of a kite at once.
If A− is of (3.2) and A+ is of (3.4), then`1 intersects `2 in two points, hence
`1=`2, which is a contradiction.
If A− is of (3.3) and A+ is of (3.3), then the kite K mentioned in (3.3) is symmetric to`1and`2, hence`1and`2are the diagonals ofK, that implies`1⊥`2, which is a contradiction.
If A− is of (3.3) and A+ is of (3.4), then`1 intersects `2 in two points, hence
`1=`2, which is a contradiction.
IfA− is of (3.4) andA+ is of (3.4), then`1=`2, which is a contradiction.
In sum, the only case with no contradiction to the conditions is the case that A− is of (3.1) andA+is of (3.2), in which caseA− is`1 and a circleC, andA+ is
`2 and a straight line`⊥2 orthogonal to`2passing through the centre ofC.
But the pointsR1, R2, R3 are collinear and they are outside of`1∪`2∪`⊥1 ∪`⊥2, hence they can not be covered by`1,C,`2 and`⊥2. This proves the theorem.
To describe the sets of injectivity is a very different task. A first step to have such decsription is the following improvement of [4, Lemma 2.1, Lemma 3.2].
Theorem 3.2. Assume that two closed segments subtend equal angles at the points of the border ∂Dof an open domainD.
(1) If the straight lines of the segments do not intersectD, they coincide.
(2) If the segments do not end on∂D, then one of the followings is valid:
(2a) the segments are collinear with two-two endpoints inside and outside ofD, and∂Dis a circle or a straight line;
(2b) D is a half plane and the segments are the reflections of each other with respect to∂D;
(2c) D is an equilateral quadrant and the segments are the reflections of each other with respect to the straight lines of the border of∂D.
Proof. Denote the segments byABandCD. LetνAB andνCDbe the visual angle functions of them, respectively, and letν=νAB−νCD. By the conditionsνvanishes on the border∂DofD.
(1) Since the straight lines of the segments do not intersect ∂D, νAB andνCD are harmonic functions on D (see Lemma A.1 in the Appendix), hence ν is also harmonic onD. Asνvanishes on∂D, the maximum principle of harmonic functions implies that ν vanishes on the whole domain D. Now Theorem 3.1 implies the coincidence of the segments by choosing suitable points inD.
(2) We have{A, B, C, D} ∩∂D=∅.
IfAB=CD, then (2a) follows from Theorem 2.4, therefore we assumeAB6≡CD from now on.
Knowing (1) we conclude that one of the straight lines AB and CD, say AB, intersects ∂D in at least one point P. Then the straight line CD also passes throughP, because νCD(P) =νAB(P) = 0. As AB6≡CDwe inferP =M0.
SinceAB6≡CD, the equiopticEis the union of subarcs of the Apollonian curves A+ andA− of the segmentsAB andCD, and therefore
(AB∪CD)∩∂D ⊆(AB∪CD)∩ E ⊆(AB∪CD)∩(A+∪ A−)⊆ {A, B, C, D, M0}.
As also∂D ∩ {A, B, C, D}=∅, we obtain (AB∪CD)∩∂D={M0}.
The straight linesAB andCD divides the plane into four open quadrantsQ−2, Q−1, Q1 and Q2, where the quadrants are indexed so that ∂Qi∩∂Q−i = {M0} (i= 1,2). It clearly follows then that∂D ⊂ {M0} ∪ Q, whereQis the union of two quadrants. These two quadrants are adjacent, if there is a neighbourhood ofM0 in which∂Dis covered by only one Apollonian curve, and they are opposite, if in any neighbourhood ofM0both Apollonian curves are necessary to cover∂D.
The pointM0is simultaneously inner or external for both segmentsABandCD, because otherwise the arcs ofA+ andA−nearM0∈ Ewould be on the compoptic
and therefore∂Dcould not passM0.
According to Observation 2.2 the intersection of an Apollonian curve wit a quadrant Qi (i = ±1,±2) is the intersection of the quadrant Qi with either the equioptic or the compoptic, hence the curves ∂D ∩ Qi are covered by exactly one of the Apollonian curves for eachi=±1,±2.
One of the quadrants that covers “half” of∂D, sayQ1, has in its border (two rays`+ and`− of the straight linesCDandAB, respectively, both starting atM0) one of the pointsA, B, C, D, say it isD.
In the quadrantQ1 the Apollonian curve A(one of A±) that covers∂Dpasses through the points M0 and D and it has an asymptotic ray `+. As ∂D is the border of the open domainD, the curve∂D ∩ Q1is connected. Denoting the point of A1 := A ∩ Q1 at the infinity by A∞ we conclude that M0 and A∞ is in one component of ¯A1:=A ∩(Q1∪∂Q1) and alsoC∈A¯1.
We proceed by considering different cases.
(2.1)Assume A is reducible. Then by Theorem 2.3 the Apollonian curveA is the straight line ` := M0A∞ and a circle C centred at a point on `. As M0 ∈/ {A, B, C, D} Theorem 2.4 proves thatABis the reflection ofCDwith respect to` (we may assume that`is the bisector ofADandBC).
Then the asymptotic straight line `⊥ of the other Apollonian curve A± of the segments is perpendicular to `, and it goes through M0 if and only if AB is the reflection ofCDwith respect to`⊥, and we get the configurations described in (2b) and (2c).
If `⊥ does not pass M0, then it can not intersect A±, because otherwise by symmetry it would have two intersections with A± that counted together with its second order intersectionB∞withA±at infinity would implyA±≡`⊥by B´ezout’s theorem.
If `⊥ does not intersect the segments, then the strip between AD and BC sep- arates B∞ from M0 in Q±2, because AD and BC intersect A± in exactly three points A, D, B∞ and B, C, B∞, respectively. If `⊥ intersects the segments, then it separatesB∞ from M0. By this disconnectedness in both cases, the arc of A±
starting fromB∞can not be in the border∂DofD, hence we get the configuration described in (2b).
(2.2)AssumeAis irreducible and has a double point, i.e. it is singular [12]. Then by [12, 5.7. Theorem] the straight lines AB, CD and either AC and BD or AD andBC are tangent to the same circleC centred to the double pointS. Without loss of generality we may assume that the straight linesAB,CD,AC andBD are tangent to the circleC.
Let`be the asymptotic straight line ofAand let`kbe the straight line throughS that is parallel to`. The straight line`k intersects Aexactly in S, but the other straight lines throughSintersectAin exactly one more point. This means that the connected part of A1=A ∩ Q1that contains A∞ is in the domainF of the angle DS`k∠. This domain does not containsM0 andAintersects its border exactly in the pointD. This proves thatM0andAcannot be in one component ofA1, hence this case can not happen.
(2.3)Assume that Ais irreducible, has no double point and has only one com- ponent. As A intersects the straight line CD in three points, it has a bounded and an unbounded (asymptotic) component (arc) on both sides ofCD, and these components are separated, becauseAdoes not have double point.
EitherM0separatesDandC, orCseparatesM0andD, orDseparatesM0andC.
IfM0separates the pointsC andD, then it is on the bounded arc ofAbetween C andD, andA∞is on the unbounded components ofA1, henceM0 andA∞ can not be in a common component ofA1.
IfC (orD) separates the pointsM0 and D (C, respectively), then it is on the bounded arc ofAbetweenM0andD(C, respectively), and thereforeM0andA∞is in a common component ofA1 if the border of the quadrantQ1contains neitherC norD. This contradicts the assumption thatD∈∂Q1, henceM0 andA∞can not be in a common component ofA1.
(2.4)Assume thatAis irreducible, has no double point and consists of two com- ponents. As Aintersects the straight lineCD in three points, two of these points
are on the bounded component, and the third one is on the unbounded. As M0
andA∞are in the same component of A1, they are also in a common component ofA, hence M0 is in the unbounded component of A. Using the above reasoning, we get that the bounded component of Acontains the points A, B, C, D which is clearly a contradiction4. This contradiction shows thatM0 andA∞can not be in a common component ofA1.
The theorem is now completely proven.
Note that Theorem 3.2 fails without the condition {A, B, C, D} ∩∂D = ∅ as shown5 by the figures in Lemma 2.1. Nevertheless we obtain the following as a corollary.
Theorem 3.3. If the centres of three circles of equal radius % form a triangle so that every heights of that triangle is bigger than 2%, then the union of the circles constitute a set of injectivity.
For the proof we only have to observe, that a segment can intersect only two of the circles.
Appendix A
LetB be a convex body with the boundary∂B, and letg: [−1,1]→R2be a C2 curve outsideB, that is parametrized by arc length and has no tangents meetingB.
Denote the tangent line, the tangent vector and the curvature of the curve g at g(s), by`(s),t(s) andκ(s), respectively. LetTa(s) andTb(s) be the tangents ofB through the pointg(s) so that their respective unit directional vectorsta(s),tb(s) pointing towardBare such thatta(s) is a positive linear combination oftb(s) and t(s). Letα(s) and β(s) be the respective angles of ta(s) and tb(s) tot(s). Then 0< α(s)< β(s)< πandν(s) =β(s)−α(s) is the shadow picture ofBatg(s).
Let A(s) = B ∩ Ta(s) and B(s) = B ∩ Tb(s) and define the points A±(s) :=
limx&0A(s±x),B±(s) := limx&0B(s±x). Their respective distances from g(s) area±(s) =|A±(s)−g(s)|andb±(s) =|B±(s)−g(s)|.
IfA+(s) = A−(s) and B+(s) =B−(s), then letA(s) = A−(s), B(s) =B−(s), a(s) =a−(s),b(s) =b−(s) and let κa(s), κb(s) be the (maybe infinite) curvatures of∂BatA(s) andB(s), respectively.
If more curves occur, then they and their objects are indexed consequently. A function without argument is understood at the appropriate parameters∈[−1,1].
Lemma A.1. Let B be a convex body and letΣB denote the union of the straight lines of the segments in∂B.
(1) The visual angle function ofB is subharmonic inR2\ B \ΣB.
(2) The visual angle function of B is harmonic at a point P ∈R2\ B \ΣB if and only if every tangent of Bthrough P touchesB in a singular point.
Proof. LetP∈R2\B \ΣBand take two orthogonal straight linesg1andg2through P so thatg1(0) =g2(0) =P. ThenA+i (0) =A−i (0) andB+i (0) =Bi−(0) (i= 1,2),
4Using the main involution [12, 4.6.] and [12, 4.3. Theorem] one sees{A, B}∗={CD}(and {A, B}={CD}∗), and by [12, 4.19.] this implies that the pointsA, B, C, D can not be in the same component.
5Look for the points nearA and C on the first and second figures, or the points near the segmentABon the third figure.
and therefore the second equation in [6, Lemma 1] implies
∆ν(P) = ¨ν1(0) + ¨ν2(0)
= sin 2β1
b2 −sin 2α1
a2 +sin2β1
b3κb +sin2α1
a3κa + +sin(2β1+π)
b2 −sin(2α1+π)
a2 +sin2(β1+π/2)
b3κb +sin2(α1+π/2) a3κa
= 1
b3κb + 1 a3κa.
This proves statement (1), because b, a, κb and κa are nonnegative functions.
Statement (2) is justified because κb =κa = +∞ if and only if the tangents goes
through singular points ofB.
References
[1] Brieskorn, E. and Kn¨orrer, H.,Plane algebraic curves, Birkh¨auser Verlag, Basel, 1986.
[2] Gardner, R. J.,Geometric tomography, Encyclopedia of Math. and its Appl.58, Cambridge University Press, Cambridge, 2006 (first edition in 1996).
[3] Kincses, J., The determination of a convex set from its angle function, Discrete Comput.
Geom., 30(2003), 287–297.
[4] Kincses, J. and Kurusa, ´A., Can you recognize the shape of a figure from its shadows?, Beitr¨age zur Alg. und Geom., 36(1995), 25–34.
[5] Kincses, J., An example of a stable, even order Quadrangle which is determined by its angle function, Discrete Geometry, in honor of W. Kuperberg’s 60th birthday(ed.: A. Bezdek), CRC Press (Marcel Dekker), New York – Basel, 2003, 367–372.
[6] Kurusa, ´A., You can recognize the shape of a figure by its shadows!, Geom. Dedicata, 59 (1996), 103–112.
[7] Kurusa, ´A., The shadow picture problem for nonintersecting curves, Geom. Dedicata, 59 (1996), 113–125.
[8] Kurusa, ´A., Is a convex plane body determined by an isoptic?, Beitr¨age Algebra Geom., 53 (2012), 281–294; DOI: 10.1007/s13366-011-0074-2.
[9] Kurusa, ´A., Equioptics of segments: generalizing Apollonius’ theorem, Polygon, 21(2013), 43–57 (in hungarian: “Szakaszok ekvioptikusai: Apoll´oniosz t´etel´enek ´altal´anos´ıt´asa”).
[10] Kurusa, ´A., Visual distinguishability of polygons, Beitr¨age Algebra Geom. (2013), DOI:
10.1007/s13366-012-0121-7.
[11] Pamfilos, P. and Thoma, A., Apollonian cubics: An application of group theory to a problem in Euclidean geometry, Mathematics Magazine, 72(1999), 356–366.
[12] Pamfilos, P.,Theory of Isoptic cubics, Help file of Isoptikon program that is freely available at http://www.math.uoc.gr/∼pamfilos/#iso , 1998.
Bolyai Institute, University of Szeged, Aradi v´ertan´uk tere 1., H-6720 Szeged, Hun- gary
E-mail address:kurusa@math.u-szeged.hu
